/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Find the linearization of \(f(x)... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 2

Find the linearization of \(f(x)\) at \(x=a\). \(f(x)=x^{3}, a=2\)

Short Answer

Expert verified
The linearization of \(f(x)=x^{3}\) at \(x=2\) is \(L(x) = 12x - 16\).

Step by step solution

01

Identify the function and point of linearization

The function to be linearized is given as \(f(x)=x^{3}\) and the point of linearization is \(x=a\) where \(a=2\).
02

Compute the function value at \(x=a\)

Evaluate the function at the point \(x=2\): \[ f(2) = 2^{3} = 8 \]
03

Compute the derivative of the function

The derivative of \(f(x) = x^3\) is \[ f'(x) = 3x^2 \]
04

Evaluate the derivative at \(x=a\)

Evaluate the derivative at the point \(x=2\):\[ f'(2) = 3(2^2) = 3 \times 4 = 12 \]
05

Construct the linearization L(x)

The linearization of the function \(f(x)\) at \(x=a\) is given by the formula: \[ L(x) = f(a) + f'(a)(x - a) \]Substitute \(f(2) = 8\), \(f'(2) = 12\), and \(a = 2\):\[ L(x) = 8 + 12(x - 2) \]Simplify the expression:\[ L(x) = 12x - 16\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Function
Understanding the derivative of a function is crucial to grasping linearization. The derivative represents the rate at which the function changes with respect to its variable. For a function like \(f(x) = x^3\), the derivative \(f'(x)\) can be found using the power rule. The rule states that if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). Applying this, the derivative of \(x^3\) is \(3x^2\). This derivative helps us understand how the function behaves near the point of linearization.
Compute Function Value
The next step in linearization is to compute the function value at a specific point, \(x=a\). For instance, in our given problem, we have \(f(x) = x^3\) and we are interested in the point \(x=2\). Plugging this value into the function, we get \(f(2) = 2^3 = 8\). This calculation gives us the function's value at the given point, which is essential for the linear approximation.
Linear Approximation
Linear approximation, or linearization, involves creating a linear formula to estimate the value of the function near a specific point. The general formula for linearization around \(x=a\) is \(L(x) = f(a) + f'(a)(x - a)\). Using our previous calculations, we know \(f(2) = 8\) and \(f'(2) = 12\). By substituting these values into the formula, we get \(L(x) = 8 + 12(x - 2)\). Simplifying this leads to \(L(x) = 12x - 16\). This linear function approximates \(f(x)\) near \(x=2\).
Evaluation of Derivatives
Evaluating the derivative at specific points provides crucial information about the function's behavior. For \(f(x) = x^3\), the derivative at \(x=2\) is calculated as \(f'(2) = 3(2^2) = 12\). These evaluations help us understand how steep or flat the function is at \(x=2\). In linearization, they directly contribute to forming the equation \(L(x)\) that approximates the function near the point. This is especially useful in various applications, as it allows us to make quick estimates without delving deeply into the complex behavior of the original function.

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Most popular questions from this chapter

Differentiate implicily to find \(d y / d x\). Then find the slope of the curve at the given point. $$ 4 x^{3}-y^{4}-3 y+5 x+1=0 ; \quad(1,-2) $$

Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. When no interval is specified, use the real line \((-\infty, \infty)\). $$ f(x)=\frac{1}{x-2 \sin x} ;(0, \pi / 2) $$

Differentiate implicily to find \(d y / d x\). $$ x^{2}+y^{2}=25 $$

Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. When no interval is specified, use the real line \((-\infty, \infty)\). $$ f(x)=\tan x-2 \sec x, \quad(\pi / 2,3 \pi / 2) $$

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